Simple Modular equation

Question

Let $s,t,n$ be 3 non-zero positive integers. We set $s+1=nt$. If $n$ is odd,find $n$ such that: $$s \equiv 1 \pmod 3$$ $$t\equiv 1 \pmod 3$$ I know the answer is very likely simple. I just suck at congruence equation. Any input will be greatly appreciated. Thanks.

Answer 1

$t=1+3m$ and $s=1+3q$ for arbitrary $m,q$.

$s+1=2+3q=nt=n+3nm$. Hence: $n = \frac{2+3q}{1+3m}$.

$n$ must be a natural number and so $1+3m$ must divide $2+3q$. This is e.g. satisfied for all pairs $(q,m)=(2m,m)$. But $n$ must be odd!

Hence the pairs $(q,m) = (1+5m,m)$ can be tried. More General, all pairs $(q,m)=(p+(2+3p)m,m)$ are possible.

Answer 2

$s$ and $t$ are solutions of the system

$s\equiv 1(3)$

$t\equiv 1 (3)$

$\frac{s+1}{t}=n(n \text{odd})$

$\Leftrightarrow n\equiv 5(6)$

Proof: $n$ odd means $n\equiv 1(2)$ $s+1=tn$ implies $n\equiv 2(3)$

so $n\equiv 5(6)$

conversely, if $n\equiv 5(6)$ then

$s=n-1$ and $t=1$ is a solution of the system